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THERMOCHEMISTRY
Studies the quantity of heat energy released or absorbed in a chemical
reaction.
Ex: the burning of fuel: is a heat-evolving reaction
Heat :
is a form of energy
Energy: -
the potential to do work (to move matter)
exists in many different forms:
- Electrical energy
- Kinetic Energy (energy of motion)
- Light energy
- Heat energy
- Chemical energy (energy of substances)
Different forms of energy can be interconverted
Ex: Light bulb:
electrical energy is converted to: - light energy and
- heat energy
Car engine:
chemical energy
(from gasoline)
heat energy
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kinetic energy
Definition:
KINETIC ENERGY (Ek)
is the energy associated with a moving object
Ek = ½ mv2 , where
NOTE:
m = mass of the moving object
(in kg)
v = speed of the moving object
(in m/s)
Two objects moving at different speeds may have the same
kinetic energy
Ex: slow moving truck and fast moving sports’ car
SI unit of energy = from formula of energy = kg . m2/s2 = joule (J)
The joule (J) is a very a very small unit
Non-SI unit of heat energy : The calorie (cal)
amount of energy required to raise the
temperature of one gram of water by one
degree Celsius
1 cal = 4.184 J
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Definition:
Example:
POTENTIAL ENERGY (Ep)
- the energy of an object has because of its position in a
field of force
- is stored energy
Any object above ground level possesses Potential Energy
since it is attracted by the gravitational force. As such, it has
the potential to fall, and therefore to move.
(Potential Energy changes into Kinetic Energy)
Formula: Ep = mgh,
where
m = mass of the object
g = constant acceleration of gravity
h = height (level) of the object above
ground level
Note: differences of level are usually more important
As an object falls toward ground level:
- its Ep decreases (h decreases)
- its Ek increases ( object speeds up; v increases)
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Definition:
INTERNAL ENERGY (U)
the sum oft he kinetic and potential energies of the particles
making up a substance.
In general: Total Energy of a Substance = U = Ek + Ep
NOTE:
A substance at rest, in a flask in the laboratory
- possesses no kinetic energy
- potential energy is ignored
- possesses Internal Energy
Total Energy of the Substance = Internal Energy of the Substance = U
Conversions of different forms of energy are governed by :
The Law of Conservation of Energy:
Energy may be converted from one form to another, but the total
quantity of energy remains constant.
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HEAT OF REACTION
Thermodynamics: is the science of the relationships between heat energy
and other forms of energy
Ek , Ep , U
: are thermodynamic properties of matter
Thermodynamic System:
The Surroundings:
the substance or mixture of substance
undergoing a physical or a chemical change
everything around the thermodynamic system
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Definition of Heat:
The energy that flows into, or out of a system because of a difference in
temperature between the thermodynamic system and its surroundings.
This assumes that the system and its surroundings are in thermal contact,
(that is, they are not thermally insulated).
Example:
hot coffee in a thermos is thermally insulated from its
surroundings
Direction of Heat Flow :
Heat always flows from a region of higher temperature to a region of
lower temperature until the temperatures of the two regions become
equal (until thermal equilibrium is established)
1.
If heat energy flows into a system from its surroundings, its internal
energy increases:
Ex: A glass of ice water slow reaches room temperature.
The internal energy (U) of the water increases, as it warms up.
2.
If heat energy flows out from a system into the surroundings, its
internal energy decreases
Ex: A cup of hot tea slowly reaches room temperature
The internal energy (U) of the tea decreases, as it cools down.
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EXPLANATION OF HEAT FLOW
HOTTER GAS
FASTER MOVING MOLECULES
Average Molecular Speed
of Hotter Gas
Faster moving molecules slow down
COLDER GAS
SLOWER MOVING MOLECULES
Average Molecular Speed
of Colder Gas
Slower moving molecules speed up
The Average Molecular Speed in the two containers become equal.
The temperature of the two gases become equal.
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Heat and Temperature
A given quantity of heat will raise the temperature of a gas more if the
sample is small.
Reason: The Absolute Temperature is directly proportional to the
Average Kinetic Energy (Ek).
- When you add heat to a gas, you increase its total Ek
- The increase in Ek will be distributed over all the
molecules in the sample
- The increase in average Ek per individual molecule (the
increase in temperature) will be greater if there are fewer
molecules.
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Symbol for Heat: q
In a chemical equation, q is always considered a reactant
heat is absorbed by the system
heat is evolved by the system
from surroundings
to the surroundings
Reactants + q  Products
Reactants  Products + q
By convention, in a chemical equation q (the Heat term) is always considered a reactant.
Reactants + q  Products
Reactants - q  Products
q is positive
q is negative
Temperature of the surroundings decreases Temperature of the surroundings increases
Reaction vessel cools
Reaction vessel warms
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Sample Problem
1. Ammonia burns in the presence of a platinum catalyst to produce
nitrogen monoxide.
4NH3(g)
+
5O2(g)
 4NO(g)
+
6H2O(l)
In an experiment, 4 mol of NH3 is burned and evolves 1170 kJ of heat.
Is the reaction endothermic or exothermic ?
Answer: exothermic (heat is evolved)
What is the value of q ?
Answer: q = - 1170 kJ
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ENTHALPY AND ENTHALPY CHANGE
Enthalpy is an extensive property of a substance that can be used to
obtain the heat absorbed or evolved in a chemical reaction
(extensive property = property that depends on the amount of substance)
Symbol for Enthalpy: H
Enthalpy (H) is a State Function:
-
A State Function is a property of a system that depends only on its
present state and independent of any previous history of the system
Ex: Hiking in the mountains
You may take several
alternative routes to reach a
certain peak
altitude reached
distance traveled
is analogous to a state function is not analogous to a state function
(difference in altitude = 3600 ft)
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ENTHALPY OF REACTION:
Consider a chemical reaction:
REACTANTS
PRODUCTS
At first:
Reactants only
No Products
The Enthalpy of the system =
The Enthalpy of the Reactants =
H (reactants)
At the end:
REACTANTS
No Reactants
The Enthalpy of the system =
PRODUCTS
Products only
The Enthalpy of the Products =
H (products)
The change in enthalpy = H = Enthalpy of Reaction = H(products) - H(reactants)
The value of H: - is independent of the details of the reaction
- depends only on the initial state (reactants) and the final state (products)
Recall that H is a state function!
THE ENTHALPY OF REACTION EQUALS THE HEAT OF REACTION AT CONSTANT
PRESSURE:
H = qp where:
H = Enthalpy of reaction
qp = Heat of reaction at constant pressure
OMIT: ENTHALPY AND INTERNAL ENERGY (pages 240—241)
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THERMOCHEMICAL EQUATIONS
The chemical equation for a reaction (including state labels) in which:
the equation is given a molar interpretation.
the Enthalpy Change (H) of reaction is written directly after the equation
Example:
2H2(g)
2 mol of
H2 gas
+
O2(g)
reacts with
1 mole
O2
H = - 483.7 kJ
2H2O(g)
to produce
2 mol of
water
vapor
and
483.7 kJ evolves
NOTE: Indicating phase labels in thermochemical equations is very important
Consider:
2H2(g) + O2(g)
2H2O(l)
H = - 571.7 kJ
This equation says:
2 mol of hydrogen gas reacts with one mole of oxygen gas to produce
2 mole of liquid water, and 571.7 kJ of heat evolves.
(additional heat is released when water vapor condenses to liquid)
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1.
MANIPULATING THERMOCHEMICAL EQUATIONS
Changing the coefficients of the balanced chemical equation, changes
the value of H accordingly
Example: Liquid carbon disulfide burns in air, producing carbon dioxide
gas and sulfur dioxide gas.
1 CS2(l) + 3 O2(g)
1 CO2(g) + 2SO2(g) H = -1075 kJ
What is H for the following equation ?
½ CS2(l)
Note:
+
3/2 O2(g)
½ CO2(g) + 1 SO2(g)
H = ?
All the coefficients are halved
1075 kJ
Therefore: H = -  = - 537.5 kJ
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RULE 1
When a thermochemical equation is multiplied (or divided) by any
factor, the value of H for the new equation is obtained by
multiplying(or dividing) the value of H in the original equation by
that same factor.
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2.
Reversing the thermochemical equation, changes the sign of H
accordingly
Example:
2H2(g) + O2(g)
2H2O(l)
H = - 571.7 kJ (exothermic)
2H2O(l)
2H2(g)
H = + 571.7 kJ(endothermic)
+ O2(g)
RULE 2
When a thermochemical equation is reversed, the value of H is
reversed in sign
Sample Problem
With a platinum catalyst, ammonia gas will burn in oxygen gas to give
gaseous nitrogen monoxide and water vapor:
4 NH3(g)
+
5 O2(g)
4 NO(g)
+
6 H2O(g)
H = - 906 kJ
What is the enthalpy change for the following reaction?
NO(g)
+
3/2 H2O(g)
NH3(g)
+
5/4 O2(g) H = ?
Step 1: First reverse the given equation:
4 NO(g)
+
6 H2O(g)
4 NH3(g) + 5 O2(g)
H = + 906 kJ
Step 2: Divide the coefficients and H by 4:
NO(g)
+
3/2 H2O(g)
NH3(g) + 5/4 O2(g)
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H = + 227 kJ
APPLYING STOICHIOMETRY TO HEATS OF REACTION
Sample problem:
Propane, C3H8, is a common fuel gas. Use the following to calculate the
grams of propane you would need to provide 255 kJ of heat.
C3H8 + 5 O2(g)
?g
3 CO2(g) + 4 H2O(g)
1 mol C3H8
44.06 g
? g C3H8 = 255 kJ x  x  =
2044 kJ
1 mol C3H8
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H = -2044 kJ
H = - 255 kJ
5.50 g
HEAT CAPACITY AND SPECIFIC HEAT
HEAT CAPACITY , C
The quantity of heat needed to raise the temperature of a sample of a
substance one degree Celsius (or one degree Kelvin)
Recall : K = 0C + 273
NOTE:
K
(Change in
temperature
in Kelvin)
q = C t where
=
 0C
Change in
temperature
in 0C)
q = quantity of heat required to produce
the temperature change
t
C
= t f - ti
= change in temperature from an initial
temperature, ti,to a final temperature, tf
= Heat capacity
HEAT CAPACITY(C) is directly proportional to the mass of
substance
MOLAR HEAT CAPACITY
Heat capacity for one mole of substance
SPECIFIC HEAT CAPACITY or SPECIFIC HEAT, s
Quantity of heat required to raise the temperature of one gram of substance
by one degree Celsius (or one Kelvin) at constant pressure
q = s x m x t
where
s = Specific Heat
m = mass
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t = temperature difference
Specific Heat and Molar Heat Capacity:
- are different for different substances
- are physical constants, characteristic for a specific substance
- are slightly temperature dependent (usually given at 250 C)
- are given in Tables (Table 6.1, page 246)
cal
J
Specific Heat of water = 1.00  = 4.18 
g x 0C
g x 0C
Sample Problem
How much heat (in joules) must be used to raise the temperature of 185 g
of water from 15 0C to 96 0C ?
M = 185 g
ti = 15 0C
tf = 96 0C
t = 96 0C - 15 0C = 81 0C
q = s x m x t
J
q =( 4.18 ) (185 g) ( 810C )
g x 0C
J
s = 4.18 
g x 0C
q = 6.3 x 104 J = 63 kJ
q=?
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EXPERIMENTAL MEASUREMENT OF HEAT OF REACTION
Calorimeter
a device used to measure the heat absorbed or evolved during a
physical or chemical change
is an insulated container (styrofoam coffee cup or a Snack-Jar
equipped with a thermometer
used to measure temperature changes which are then related to
changes in heat energy
Sample Problem
50.0 mL each of 1.0 M HCl and 1.0 M NaOH, at room temperature (20.0 O
C) are mixed. The temperature of the resulting NaCl solution increases to
27.5 O C.
The density of the resulting NaCl solution is 1.02 g/mL.
The specific heat of the resulting NaCl solution is 4.06 J/g 0 C
Calculate the Heat of Neutralization of HCl(aq) and NaOH(aq) in
KJ/mol NaCl produced
50.0 mL HCl, 1.0 M (20.00C)
50.0 mL NaOH, 1.0 M (20.00C)
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HCl(aq) +
NaOH(aq)
NaCl(aq) + H2O(l)
+
Heat of Neutralization (H)
Heat given off by the Neutralization Rxn = Heat gained by the NaCl solution
HCl(aq) + NaOH(aq) - (H)
NaCl(aq) + H2O(l) (exothermic)
Mass of solution: V x D = (100.0 mL) x (1.02 g/mL)=
m
= 102 g
Specific heat = 4.06 J/g 0 C
t
= 7.5 0 C
H =
H =
102 g
- s x m x t
- (4.06 J/g 0 C) (102 g) (7.5 0 C) = - 3100 J = - 3.1 kJ (exothermic reaction)
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3.1 kJ
?  = ?
mol of NaCl produced
HCl(aq)
+
NaOH(aq) 50.0 mL 1.0 M
50.0 mL 1.0 M
Note:
Therefore:
? mol NaCl produced = ?
(H)
NaCl(aq) + H2O(l) (exothermic)
? mol
Reactants are in stoichiometric ratio (No Limiting Reactant, No Reactant in excess)
The resulting moles of NaCl can be calculated from the volumes of either NaOH
solution or HCl solution.
1.0 mol HCl
1 mol NaCl
? mol NaCl = 50.0mL HCl x  x 
1000 mL HCl
1 mol HCl
3.1 KJ
H = - 
0.050 mol NaCl produced
= - 62 KJ/mol
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= 0.0050 mol NaCl
Sample Problem
2.08 g of NaOH are dissolved in 50.0 mL of water whose temperature is 20.0 O C. The
temperature of the NaOH solution after mixing increases to 28.8 0 C.
The specific heat of the resulting NaOH solution is about 3.93 J/g 0 C
Calculate the Heat of Solution of NaOH in KJ/mol
NaOH(s)
2.08 g
+
H2O(l) 50.0 g
H
NaOH(aq) (exothermic)
52.1 g
Heat of solution of NaOH = H = Heat gained by the NaOH solution
Mass of solution
= 52.1 g
Initial Temperature
= 20.0 O C.
= - 1.8 KJ
Final Temperature
= 28.8 0 C
t
= 8.8 0 C
NaOH
Specific heat = 3.93 J/g 0 C
kJ
Heat of solution = H = ? 
mol NaOH
H = s x m x t
H =
- (3.93 J/g 0 C) (52.1 g) (8.8 0 C) = - 1800 J
- 1.8 kJ
1 mol
? mol NaOH = 2.08 g  = 0.052 mol
40.0 g
- 1.8 kJ
H =  = - 35 KJ/mol
0.052 mol NaOH
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HESS’S LAW
Hess’s Law of Summation
For a chemical equation that can be written as the sum of two or more steps, the enthalpy
change for the overall equation equals the sum of the enthalpy changes for the individual
steps.
Sample Problem
Ammonia will burn in the presence of a platinum catalyst to produce nitric oxide, NO
4NH3(g) +
5O2(g)

4NO(g) +
6H2O(g)
H = ?
What is the heat of reaction at constant pressure ?
Use the following thermochemical equations:
N2(g)
+
O2(g)
2NO(g)
H1 = + 180.6 kJ (Eq 1)
N2(g)
+
3H2(g)
2NH3(g)
H2 = - 91.8 kJ
(Eq 2)
2H2(g)
+
O2(g)
2H2O(g)
H3 = - 483.7 kJ
(Eq 3)
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4NH3(g) +
N2(g)
N2(g)
2H2(g)
Step 1:
5O2(g)
4NO(g) +
6H2O(g)
H = ?
+ O2(g)
2NO(g)
H1 = + 180.6 kJ (Eq 1)
+ 3H2(g)
2NH3(g)
H2 = - 91.8 kJ
(Eq 2)
+ O2(g)
2H2O(g)
H3 = - 483.7 kJ (Eq 3)
To get 4NH3(g) on the reactant side: Multiply Eq 2 by 2, and reverse its its direction:
N2(g)
+
4NH3(g)
2H2(g) + O2(g)
Step 2:

O2(g)
2N2(g)
+
2NO(g) H1 = +180.6 kJ
(Eq 1)
6H2(g)
H2 = ( -91.8 kJ) (-2)=+184 kJ (Eq 2)
2H2O(g)
H3 = - 483.7 kJ
(Eq 3)
To get 4NO(g) on the product side: Multiply Eq 1 by 2:
2N2(g)
+
4NH3(g)
2H2(g) + O2(g)
2O2(g)
2N2(g)
+
4NO(g)
6H2(g)
2H2O(g)
H1 =(+180.6 kJ)(2) =+361.2 kJ (Eq 1)
H2 = +184 kJ
(Eq 2)
H3 = - 483.7 kJ
(Eq 3)
Step 3: To get 6H2O(g) on the product side: Multiply Eq 3 by 3:
2N2(g)
+
4NH3(g)
6H2(g) +3O2(g)
2O2(g)
4NO(g)
2N2(g) + 6H2(g)
6H2O(g)
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H1 = +361.2 kJ
(Eq 1)
H2 = +184 kJ
(Eq 2)
H3 = - 483.7 kJ (3) = - 1450 kJ (Eq 3)
Step 4: Add the 3 steps (equations) and cancel out 2N2(g) and 6H2(g)
2N2(g)
+
4NH3(g)
6H2(g) +3O2(g)
2O2(g)
2O2(g)
4NO(g)
2N2(g) + 6H2(g)
6H2O(g)
+ 4NH3(g) +3O2(g)
H1 =
+361.2 kJ (Eq 1)
H2 =
+184 kJ(Eq 2)
H3 = - 483.7 kJ (3) = - 1450 kJ (Eq 3)
4NO(g) + 6H2O(g)
HT = H1 + H2 + H3
HT = +361.2 kJ + 184 kJ - 1450 kJ
= -905 kJ
SUM:
5O2(g)
+ 4NH3(g)
4NO(g) + 6H2O(g)
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HT =
-905 kJ
An Enthalpy diagram illustrating how Hess’s law applies to this example is
shown below.
REACTANTS
4 NH3(g) + 5 O2(g)
+ 184 kJ
- 905 kJ
- 1450 kJ
PRODUCTS
4 NO(g) + 6 H2Og)
+ 361.2 kJ
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STANDARD ENTHALPIES OF FORMATION
Standard State
= Standard Thermodynamic
Conditions
indicated by a superscript: (o)
(1 atm pressure and 25o C)
Standard Enthalpy of Reaction, Ho (Read delta H zero)
The Enthalpy change for a reaction in which reactants in their standard states
yield products in their standard state.
Tables usually give Enthalpy changes for Formation Reactions only.
Formation Reactions:
Reactions in which compounds are formed from their elements
A + B
AB
Example:
C(graphite)
+
2Cl2(g)
 CCl4(l)
Hof = -139 kJ/mol
Standard Enthalpy of formation for liquid CCl4 from graphite = 139 kJ/mol
Note:
It is important to indicate the form in which the element participates in a
formation reaction, since the value of Hof is determined by this.
Carbon may exist in other forms as well, referred to as allotropes
(ex: diamond). However, graphite is the most stable form of carbon
(Graphite is referred to as the reference form of carbon)
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Reference form of an element :
The most stable form of the element under standard thermodynamic
conditions (1 atm pressure and 25o C)
Hof of an element in its reference form = 0
For example:
Hof (graphite) = 0
Hof (diamond) = 1.9 kJ/mol
Meaning:
C(graphite)
Hof
Hof = + 1.9 kJ/mol (endothermic)
C(diamond)
=
=
STANDARD ENTHALPY OF FORMATION
STANDARD HEAT OF FORMATION
=
is the enthalpy change for the formation of one mole of
the substance in its standard state from its elements in
their reference form and in their standard states.
(For Values, see Table 6.2, page 256 and in Appendix C)
Applications:
Using Hess’s Law, the given standard enthalpies of formation can be used
to find the standard enthalpy change for a reaction.
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Sample Problem
The cooling effect of alcohol on the skin is due to its evaporation.
Calculate the heat of vaporization of ethanol (ethyl alcohol), C2H5OH.
C2H5OH(l)
Ho vaporization = ?
C2H5OH(g)
From Table 6.2 or Appendix C:
Hof [C2H5OH(l)] =
- 277.6 kJ/mol
Hof [C2H5OH(g)] =
-235.4 kJ/mol
Hof
C2H5OH(l)
- 277.6 kJ/mol
Ho vaporization =
Ho vaporization =
C2H5OH(g)
-235.4 kJ/mol
Ho vaporization = ?
Hof [C2H5OH(g)]
Hof [C2H5OH(l)]
(-235.4 kJ/mol) - (-277 kJ/mol) = + 42. 2 kJ/mol
Note: Positive sign indicates that reaction is endothermic
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IN GENERAL, you can calculate the Ho for a reaction:
Ho
=
Sum of Hof of products
Ho =
 n Hof (products)
where:
=
Sum of Hof of reactants
 m Hof (reactants)
mathematical symbol for “the sum of”
n, m, = the coefficients of the substances in the
chemical equation
Sample Problem
Iron is obtained from iron ore by reduction with carbon monoxide. The
overall reaction is:
Fe2O3(s) +
3CO(g)
2Fe(s)
+
3CO2(g)
Calculate the standard enthalpy change for this equation.
See Appendix C for data.
Hof
Fe2O3(s) +
-825.5
3CO(g)
3(-110.5)
Ho =
=
 n Hof (products)
[2(0) + 3(-395.5)]
2Fe(s)
2(0)
+
3CO2(g)
3(-393.5)
(kJ)
 m Hof (reactants)
[(-825.5) + 3(-110.5) = -23.5 kJ
FUELS - FOODS, COMMERCIAL FUELS, AND ROCKET FUELS
Study on your own: Section 6.9, pp 259-262
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